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Questions: Algebra BusinessCalculus
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Find the equation of the line perpendicular to \(\displaystyle y = x + 9\) that contains \(\displaystyle \left( 3, \ 2\right)\).
The slope of the line is \(\displaystyle m = 1\) so the perpendicular line has slope \(\displaystyle m=-1\). The equation has the form \(\displaystyle y=-1x+b\). Using the point \(\displaystyle \left( 3, \ 2\right)\) gives the equation \(\displaystyle 2=-1\left(3\right)+b\) Solving for \(\displaystyle b\) gives \(\displaystyle b = 5\). The equation of the perpendicular line is \(\displaystyle y = 5 - x\).
\begin{question}Find the equation of the line perpendicular to $y = x + 9$ that contains $\left( 3, \ 2\right)$.
\soln{9cm}{The slope of the line is $m = 1$ so the perpendicular line has slope $m=-1$. The equation has the form $y=-1x+b$. Using the point $\left( 3, \ 2\right)$ gives the equation $2=-1\left(3\right)+b$ Solving for $b$ gives $b = 5$. The equation of the perpendicular line is $y = 5 - x$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the equation of the line perpendicular to <img class="equation_image" title=" \displaystyle y = x + 9 " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20x%20%2B%209%20" alt="LaTeX: \displaystyle y = x + 9 " data-equation-content=" \displaystyle y = x + 9 " /> that contains <img class="equation_image" title=" \displaystyle \left( 3, \ 2\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%203%2C%20%5C%20%202%5Cright%29%20" alt="LaTeX: \displaystyle \left( 3, \ 2\right) " data-equation-content=" \displaystyle \left( 3, \ 2\right) " /> . </p> </p>
<p> <p>The slope of the line is <img class="equation_image" title=" \displaystyle m = 1 " src="/equation_images/%20%5Cdisplaystyle%20m%20%3D%201%20" alt="LaTeX: \displaystyle m = 1 " data-equation-content=" \displaystyle m = 1 " /> so the perpendicular line has slope <img class="equation_image" title=" \displaystyle m=-1 " src="/equation_images/%20%5Cdisplaystyle%20m%3D-1%20" alt="LaTeX: \displaystyle m=-1 " data-equation-content=" \displaystyle m=-1 " /> . The equation has the form <img class="equation_image" title=" \displaystyle y=-1x+b " src="/equation_images/%20%5Cdisplaystyle%20y%3D-1x%2Bb%20" alt="LaTeX: \displaystyle y=-1x+b " data-equation-content=" \displaystyle y=-1x+b " /> . Using the point <img class="equation_image" title=" \displaystyle \left( 3, \ 2\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%203%2C%20%5C%20%202%5Cright%29%20" alt="LaTeX: \displaystyle \left( 3, \ 2\right) " data-equation-content=" \displaystyle \left( 3, \ 2\right) " /> gives the equation <img class="equation_image" title=" \displaystyle 2=-1\left(3\right)+b " src="/equation_images/%20%5Cdisplaystyle%202%3D-1%5Cleft%283%5Cright%29%2Bb%20" alt="LaTeX: \displaystyle 2=-1\left(3\right)+b " data-equation-content=" \displaystyle 2=-1\left(3\right)+b " /> Solving for <img class="equation_image" title=" \displaystyle b " src="/equation_images/%20%5Cdisplaystyle%20b%20" alt="LaTeX: \displaystyle b " data-equation-content=" \displaystyle b " /> gives <img class="equation_image" title=" \displaystyle b = 5 " src="/equation_images/%20%5Cdisplaystyle%20b%20%3D%205%20" alt="LaTeX: \displaystyle b = 5 " data-equation-content=" \displaystyle b = 5 " /> . The equation of the perpendicular line is <img class="equation_image" title=" \displaystyle y = 5 - x " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%205%20-%20x%20" alt="LaTeX: \displaystyle y = 5 - x " data-equation-content=" \displaystyle y = 5 - x " /> . </p> </p>